Determinant/Recursively/Multilinear/Fact/Proof

Let

${\displaystyle M:={\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}\,,M':={\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}}{\text{ and }}{\tilde {M}}:={\begin{pmatrix}v_{1}\\\vdots \\v_{k-1}\\u+w\\v_{k+1}\\\vdots \\v_{n}\end{pmatrix}},}$

where we denote the entries and the matrices arising from deleting a row in an analogous way. In particular, ${\displaystyle {}u=\left(a_{k1},\,\ldots ,\,a_{kn}\right)}$ and ${\displaystyle {}w=\left(a_{k1}',\,\ldots ,\,a_{kn}'\right)}$. We prove the statement by induction over ${\displaystyle {}n}$,  For ${\displaystyle {}i\neq k}$, we have ${\displaystyle {}{\tilde {a}}_{i1}=a_{i1}=a'_{i1}}$ and

${\displaystyle {}\det {\tilde {M}}_{i}=\det M_{i}+\det M'_{i}\,}$

due to the induction hypothesis. For ${\displaystyle {}i=k}$, we have ${\displaystyle {}M_{k}=M_{k}'={\tilde {M}}_{k}}$ and ${\displaystyle {}{\tilde {a}}_{k1}=a_{k1}+a'_{k1}}$. Altogether, we get

${\displaystyle {}{\begin{aligned}\det {\tilde {M}}&=\sum _{i=1}^{n}(-1)^{i+1}{\tilde {a}}_{i1}\det {\tilde {M}}_{i}\\&=\sum _{i=1,\,i\neq k}^{n}(-1)^{i+1}a_{i1}(\det {M}_{i}+\det {M}'_{i})+(-1)^{k+1}(a_{k1}+a'_{k1})(\det {\tilde {M}}_{k})\\&=\sum _{i=1,\,i\neq k}^{n}(-1)^{i+1}a_{i1}\det {M}_{i}+\sum _{i=1,\,i\neq k}^{n}(-1)^{i+1}a_{i1}\det {M}'_{i}+(-1)^{k+1}a_{k1}\det M_{k}+(-1)^{k+1}a'_{k1}\det M_{k}\\&=\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det {M}_{i}+\sum _{i=1,\,i\neq k,\,}^{n}(-1)^{i+1}a_{i1}\det {M}'_{i}+(-1)^{k+1}a'_{k1}\det M_{k}\\&=\sum _{i=1}^{n}(-1)^{i+1}a_{i1}\det {M}_{i}+\sum _{i=1}^{n}(-1)^{i+1}a'_{i1}\det {M}'_{i}\\&=\det M+\det M'.\end{aligned}}}$

The compatibility with the scalar multiplication is proved in a similar way, see exercise.